2833. Furthest Point From Origin

Intuition

Every '_' wildcard can be assigned to either 'L' or 'R' freely, so they always contribute +1 to the furthest possible distance regardless of the other moves. The non-wildcard moves produce a net displacement of |countL - countR|. Adding all wildcards to that net displacement gives the maximum reachable point.

Algorithm

1Make a single pass over the moves string, counting 'L', 'R', and '_' characters.
2Compute net = Math.Abs(countL - countR).
3Return net + countUnderline (each wildcard extends the furthest point by 1).

Example Walkthrough

Input: moves = "_L__LL__"

1.Count: L = 3, R = 0, _ = 5.
2.Net displacement without wildcards = |3 - 0| = 3.
3.All 5 wildcards go left (dominant direction): 3 + 5 = 8.

Output: 8

Common Pitfalls

•Do not try to split wildcards between L and R — they always add fully to the dominant side.
•The answer is |L - R| + wildcards, not max(L, R) + wildcards.

2833.cs

// Approach: Count L moves, R moves, and '_' wildcards separately in one pass.
// The net displacement without wildcards is |countL - countR|. Every wildcard '_'
// can freely move in whichever direction is dominant, adding 1 to the furthest distance.
// So the answer is simply Math.Abs(countL - countR) + countUnderline.
//
// Time: O(N) — single pass over the moves string.
// Space: O(1) — three integer counters only.
public class Solution
{
    public int FurthestDistanceFromOrigin(string moves)
    {
        int countL = 0;
        int countR = 0;
        int countUnderline = 0;

        foreach (char c in moves)
        {
            if (c == 'L')
                ++countL;
            else if (c == 'R')
                ++countR;
            else // c == '_'
                ++countUnderline;
        }

        return Math.Abs(countL - countR) + countUnderline;
    }
}

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